What is the derivative of 1/2 (9t^2+t)^{-3} ?

The derivative of 1/2 (9t^2+t)^{-3} is -(3(18t+1))/(2(9t^2+t)^4)

What is the first derivative of 1/2 (9t^2+t)^{-3} ?

The first derivative of 1/2 (9t^2+t)^{-3} is -(3(18t+1))/(2(9t^2+t)^4)

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derivative of

\frac{1}{2} (9 t^{2} + t)^{- 3}

- \frac{3 ( 18 t + 1 )}{2 ( 9 t ^{2} + t ) ^{4}}

Solution steps

\frac{d}{d t} (\frac{1}{2} (9 t^{2} + t)^{- 3})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{2} \frac{d}{d t} ((9 t^{2} + t)^{- 3})

Apply the chain rule:

- \frac{3}{( 9 t ^{2} + t ) ^{4}} \frac{d}{d t} (9 t^{2} + t)

= - \frac{3}{( 9 t ^{2} + t ) ^{4}} \frac{d}{d t} (9 t^{2} + t)

\frac{d}{d t} (9 t^{2} + t) = 18 t + 1

= \frac{1}{2} (- \frac{3}{( 9 t ^{2} + t ) ^{4}} (18 t + 1))

Simplify

\frac{1}{2} (- \frac{3}{( 9 t ^{2} + t ) ^{4}} (18 t + 1)) : - \frac{3 ( 18 t + 1 )}{2 ( 9 t ^{2} + t ) ^{4}}

= - \frac{3 ( 18 t + 1 )}{2 ( 9 t ^{2} + t ) ^{4}}

\int_{0}^{1} x^{2} + x d x

y^{^{''}} + y = 0, y (0) = 0

(5, 2), (- 2, - 2)

\frac{d}{d r} (4 - r^{2})

\int x (3 sec^{2} (x) + 2 tan^{2} (x)) d x

What is the derivative of 1/2 (9t^2+t)^{-3} ?
The derivative of 1/2 (9t^2+t)^{-3} is -(3(18t+1))/(2(9t^2+t)^4)
What is the first derivative of 1/2 (9t^2+t)^{-3} ?
The first derivative of 1/2 (9t^2+t)^{-3} is -(3(18t+1))/(2(9t^2+t)^4)